Expresa en forma de una sola potencia:
a) (x/2)^1/3×(x/2)^3/5
b) (-3/4)^4×(-3/4)^-2/5×(-3/4)
c) [(1+√2)^3]^3/5 ÷ (1+√2)^-1/2
d)(-1/5)^-7 ÷ (-1/5)^-7
Respuestas a la pregunta
Contestado por
202
Resolver.
Aplicas
a^n x a^m = a^(n + m)
(x/2)¹/³ x (x/2)³/⁵ =
(x/2)¹/³ ⁺ ³/⁵ = 1/3 + 3/5 = (5 + 9)/15 = 14/15
(x/2)¹⁴/¹⁵
-------------------------------------------------------------------------------
(-3/4)⁴ x (- 3/4)⁻²/⁵ x (-3/4)¹
(-3/4)⁴ ⁻²/⁵⁺¹ = 4 - 2/5 + 1=(20 - 2 + 5)/5 = 23/5
(-3/4)²³/⁵
-------------------------------------------------------------------------------
[(1 + √2)³]³/⁵ ÷ (1 + √2)⁻¹/² =
(1 - √2)⁹/⁵ ÷ (1 - √2)⁻¹/² = Aplicas a/b = a . b⁻¹
(1 + √2) ⁹/⁵ x (1 + √2)¹/² =
(1 + √2)⁹/⁵ ⁺ ¹/² = 9/5 + 1/2 =(18 + 5)/10 = 23/10
(1 + √2)²³/¹°
-------------------------------------------------------------------------------------
(- 1/5)⁻⁷ ÷ (- 1/5)⁻⁷ =
(-1/5)⁻⁷ x (-1/5)⁷ =
(-1/5)⁻⁷⁺⁷ =
(-1/5)° =
1
Aplicas
a^n x a^m = a^(n + m)
(x/2)¹/³ x (x/2)³/⁵ =
(x/2)¹/³ ⁺ ³/⁵ = 1/3 + 3/5 = (5 + 9)/15 = 14/15
(x/2)¹⁴/¹⁵
-------------------------------------------------------------------------------
(-3/4)⁴ x (- 3/4)⁻²/⁵ x (-3/4)¹
(-3/4)⁴ ⁻²/⁵⁺¹ = 4 - 2/5 + 1=(20 - 2 + 5)/5 = 23/5
(-3/4)²³/⁵
-------------------------------------------------------------------------------
[(1 + √2)³]³/⁵ ÷ (1 + √2)⁻¹/² =
(1 - √2)⁹/⁵ ÷ (1 - √2)⁻¹/² = Aplicas a/b = a . b⁻¹
(1 + √2) ⁹/⁵ x (1 + √2)¹/² =
(1 + √2)⁹/⁵ ⁺ ¹/² = 9/5 + 1/2 =(18 + 5)/10 = 23/10
(1 + √2)²³/¹°
-------------------------------------------------------------------------------------
(- 1/5)⁻⁷ ÷ (- 1/5)⁻⁷ =
(-1/5)⁻⁷ x (-1/5)⁷ =
(-1/5)⁻⁷⁺⁷ =
(-1/5)° =
1
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